Normal Approximation to Binomial Interval Probabilities

Approximating a Binomial Interval

A bounded count question asks for the probability that a binomial random variable falls between two specified integers. For example, a system might process a fixed number of independent transactions, each with the same probability of an error, and ask for the probability of between 18 and 27 errors, inclusive. When a normal approximation is appropriate, the key is to approximate the whole interval, not just one endpoint or one tail.

The binomial model and its approximation conditions are assumed to have been checked. In particular, the trials must be independent, each trial must have the same success probability, and the expected numbers of successes and failures should be large enough for a normal approximation to be reasonable. Rules of thumb for these conditions are not guarantees of accuracy; use the condition-checking material when deciding whether the approximation is suitable.

Definition: If $X\sim\operatorname{Bin}(n,p)$, then $X$ counts successes in $n$ independent trials with common success probability $p$. Write $q=1-p$. Its mean and variance are $\mu=np$ and $\sigma^2=npq$, so its standard deviation is $\sigma=\sqrt{npq}$. For an inclusive integer interval $a\leq X\leq b$, the continuity-corrected normal approximation is $$ P(a\leq X\leq b)\ \approx\ \Phi\left(\frac{b+0.5-\mu}{\sigma}\right)-\Phi\left(\frac{a-0.5-\mu}{\sigma}\right), $$ where $\Phi(z)$ is the cumulative distribution function of a standard normal random variable.

Why Both Endpoints Move by One Half

A binomial variable takes integer values, while a normal variable varies continuously. To approximate the probability of an integer count, associate each integer $k$ with the interval from $k-0.5$ to $k+0.5$. The normal area over that interval represents the probability mass at $k$. This is the continuity correction: it makes the continuous interval cover the same integer counts as the discrete event.

For the event $a\leq X\leq b$, the included integer counts are $a,a+1,\ldots,b$. Their associated unit-width intervals join together from $a-0.5$ through $b+0.5$. Therefore the normal probability is taken between $a-0.5$ and $b+0.5$, not between $a$ and $b$.

\begin{aligned} P(a\leq X\leq b) &=\sum_{k=a}^{b}P(X=k)\\ &\approx \sum_{k=a}^{b}P(k-0.5<N<k+0.5)\\ &=P(a-0.5<N<b+0.5), \end{aligned}

Here $N$ denotes a normal random variable with mean $\mu=np$ and variance $\sigma^2=npq$. The approximation in the second line replaces each binomial point probability by the normal area over the half-unit cell around that count. Standardizing the final interval gives the formula in the definition. The normal distribution assigns probability zero to a single point, so whether the continuous inequalities are strict or inclusive does not change the probability.

Specify the binomial model.
Identify $n$, $p$, and the inclusive integer bounds $a$ and $b$. Confirm that the event is genuinely $a\leq X\leq b$.

Calculate the normal parameters.
Use $\mu=np$ and $\sigma=\sqrt{np(1-p)}$. Keep the variance and standard deviation distinct.

Correct both boundaries.
Replace the lower integer bound by $a-0.5$ and the upper integer bound by $b+0.5$.

Standardize and subtract.
Find the upper cumulative probability minus the lower cumulative probability. Round the final probability to four decimal places.

Worked Interval Calculations

Worked Example: A Central Interval of Quality Checks

A hypothetical inspection process checks 200 items. Each item independently fails inspection with probability $0.40$. Let $X$ be the number of failures. Approximate the probability of between 72 and 88 failures, inclusive.

Method and model. The count has the binomial model $X\sim\operatorname{Bin}(200,0.40)$. Thus $n=200$, $p=0.40$, $q=1-0.40=0.60$, and the requested event is $72\leq X\leq88$. The expected counts are $np=80$ failures and $nq=120$ nonfailures.

Calculate the normal parameters.

\mu=np=200(0.40)=80,\qquad \sigma^2=npq=200(0.40)(0.60)=48,\qquad \sigma=\sqrt{48}\approx6.9282.

Apply the continuity correction. The inclusive count interval becomes $71.5<N<88.5$. The standardized endpoints are

z_L=\frac{71.5-80}{6.9282}\approx-1.23,\qquad z_U=\frac{88.5-80}{6.9282}\approx1.23.

Using standard normal cumulative probabilities, $\Phi(1.23)=0.8907$ and $\Phi(-1.23)=0.1093$. Therefore

P(72\leq X\leq88)\approx\Phi(1.23)-\Phi(-1.23) =0.8907-0.1093=0.7814.

The approximated probability is 0.7814, or about 78.14%. The result is symmetric around the mean count of 80, which is consistent with the equal-magnitude standardized endpoints. The common mistake avoided here is using 72 and 88 directly as normal boundaries; that would omit the half-unit coverage at both ends.

Worked Example: An Interval That Is Not Centered at the Mean

A hypothetical network records whether each of 400 requests requires a manual review. Each request independently requires review with probability $0.25$. Approximate the probability that the number of reviews is from 90 through 115, inclusive.

Method and model. Let $X\sim\operatorname{Bin}(400,0.25)$, so $n=400$, $p=0.25$, and $q=0.75$. The interval is $90\leq X\leq115$; it is not centered at the expected count.

Calculate the parameters.

\mu=400(0.25)=100,\qquad \sigma^2=400(0.25)(0.75)=75,\qquad \sigma=\sqrt{75}\approx8.6603.

The expected counts are 100 reviews and 300 requests that do not require review. The corrected interval is $89.5<N<115.5$, giving

z_L=\frac{89.5-100}{8.6603}\approx-1.21,\qquad z_U=\frac{115.5-100}{8.6603}\approx1.79.

Using $\Phi(1.79)=0.9633$ and $\Phi(-1.21)=0.1131$,

P(90\leq X\leq115)\approx0.9633-0.1131=0.8502.

The approximation is 0.8502. Both cumulative probabilities are needed: the lower endpoint is not zero, so using only the upper cumulative probability would include counts below 90. The method is to subtract the lower-tail area at the corrected lower endpoint from the cumulative area at the corrected upper endpoint.

Worked Example: A Bounded Count in an Operational Forecast

In a hypothetical batch of 500 independent shipments, each shipment has probability $0.30$ of arriving later than its target. Let $X$ be the number of late shipments. Approximate the probability of between 136 and 160 late shipments, inclusive.

Method and model. Here $X\sim\operatorname{Bin}(500,0.30)$, with $n=500$, $p=0.30$, and $q=0.70$. The event is $136\leq X\leq160$. The expected counts of late and on-time shipments are 150 and 350.

Calculate and standardize.

\mu=500(0.30)=150,\qquad \sigma^2=500(0.30)(0.70)=105,\qquad \sigma=\sqrt{105}\approx10.2470.

The continuity-corrected interval is $135.5<N<160.5$, so

z_L=\frac{135.5-150}{10.2470}\approx-1.42,\qquad z_U=\frac{160.5-150}{10.2470}\approx1.02.

Using $\Phi(1.02)=0.8461$ and $\Phi(-1.42)=0.0778$,

P(136\leq X\leq160)\approx0.8461-0.0778=0.7683.

The approximation is 0.7683. The interval is not symmetric around 150: it extends 14 counts below the mean and 10 above it before continuity correction. Accordingly, the standardized endpoints are not opposites, and the probability should not be calculated by doubling a single tail. A check is that the corrected bounds are each on the appropriate side of the mean and the resulting probability lies between zero and one.

Endpoint Choices and Common Traps

The half-unit adjustment depends on the event's exact inequalities. For integer-valued $X$, an event such as $a<X\leq b$ includes counts $a+1,\ldots,b$, so its corrected continuous interval is from $a+0.5$ to $b+0.5$. Likewise, $a\leq X<b$ corresponds to boundaries $a-0.5$ and $b-0.5$. First identify which integer values are included; only then move the endpoints.

A frequent error is to shift both endpoints in the same direction, or to apply $0.5$ to the standardized $z$-scores rather than to the count boundaries. The correction is made on the original count scale, before subtracting the mean and dividing by the standard deviation.

Common trap: For $a\leq X\leq b$, the incorrect calculation $\Phi((b-\mu)/\sigma)-\Phi((a-\mu)/\sigma)$ uses uncorrected endpoints. The correction is $\Phi((b+0.5-\mu)/\sigma)-\Phi((a-0.5-\mu)/\sigma)$. Also check that $\sigma=\sqrt{npq}$, not $npq$; dividing by the variance instead of the standard deviation produces incorrect standardized values.

A bounded interval may be narrow, broad, centered at the mean, or displaced from it. The calculation procedure is unchanged, but an approximation should not be mistaken for an exact binomial probability. Its quality depends on whether the binomial distribution is reasonably represented by a normal curve in the region of interest. The model conditions, the location of the interval, and the discrete-to-continuous correction all matter.

Key takeaway: For an inclusive interval $a\leq X\leq b$, use the binomial mean $np$, standard deviation $\sqrt{np(1-p)}$, and corrected count endpoints $a-0.5$ and $b+0.5$. Standardize each endpoint and subtract the lower cumulative probability from the upper one.

Check Your Understanding

For each problem, use a continuity-corrected normal approximation and round the final probability to four decimal places.

A hypothetical sensor produces an alert independently on each of 160 checks with probability $0.50$. Approximate the probability of 72 through 88 alerts, inclusive. (Answer: 0.8198)
A call center receives 300 independent contacts, each requiring a specialist with probability $0.20$. Approximate the probability that 48 through 72 contacts require a specialist. (Answer: 0.9282)
In a hypothetical sample of 240 devices, each independently needs a repair with probability $0.35$. Approximate the probability of 76 through 95 repairs, inclusive. (Answer: 0.8155)
A batch contains 500 independent components, each with probability $0.08$ of failing a test. Approximate the probability that 32 through 52 components fail. (Answer: 0.8995)
A hypothetical platform handles 180 independent transactions, each with probability $0.60$ of completing within a specified time. Approximate the probability that between 100 and 118 transactions, inclusive, complete within that time. (Answer: 0.8467)

What You'll Learn